The weight of $1 \, kg$ becomes $1/6$ on the moon. If the radius of the moon is $1.768 \times 10^6 \, m$,then the mass of the moon will be:

Write the equation of gravitational acceleration which is used for any height $h$ from the surface of the Earth.

If the radius of the Earth is $R$,then the height $h$ at which the value of $g$ becomes one-fourth is:

Assuming Earth to be a sphere of uniform density,what is the value of gravitational acceleration in a mine $100 \, km$ below the Earth's surface? (Given $R = 6400 \, km$,$g = 9.8 \, m/s^2$)

Acceleration due to gravity on the moon is $\frac{1}{6}$ of the acceleration due to gravity on the earth. If the ratio of densities of earth $(\rho_e)$ and moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,then the radius of the moon $R_m$ in terms of $R_e$ will be:

The ratio of the time periods of a simple pendulum at heights $2 R_E$ and $3 R_E$ from the surface of the earth is ($R_E$ is the radius of the earth).